feat : product-of-arrray-except-self

Author: ekgns33

product-of-array-except-self/ekgns33.java

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+/*
+input : array of integer
+output : array of integer that each element
+     is product of all the elements except itself
+constraint
+1) is the result of product also in range of integer?
+yes product of nums is guaranteed to fit 32-bit int
+2) how about zero?
+doesn't matter if the prduct result is zero
+3) is input array non-empty?
+yes. length of array is in range [2, 10^5]
+
+solution1) brute force
+
+calc all the product except current index element
+
+tc : O(n^2) sc : O(n) << for the result. when n is the length of input array
+
+solution 2) better?
+ds : array
+algo : hmmm we can reuse the product using prefix sum
+
+1) get prefix sum from left to right and vice versa : 2-O(n) loop + 2-O(n) space
+2) for i = 0 to n-1 when n is the length of input
+    get product of leftPrfex[i-1] * rightPrefix[i+1]
+       // edge : i == 0, i == n-1
+
+tc : O(n) sc : O(n)
+
+solution 3) optimal?
+can we reduce space?
+1) product of all elements. divide by current element.
+
+    > edge : what if current element is zero?
+    2) if there exists only one zero:
+        all the elements except zero index will be zero
+    3) if there exist multiple zeros:
+        all the elements are zero
+    4) if there is no zero
+        do 1)
+ */
+class Solution {
+  public int[] productExceptSelf(int[] nums) {
+    int n = nums.length, product = 1, zeroCount = 0;
+    for(int num : nums) {
+      if(num == 0) {
+        zeroCount ++;
+        if(zeroCount > 1) break;
+      } else {
+        product *= num;
+      }
+    }
+
+    int[] answer = new int[n];
+    if(zeroCount > 1) {
+      return answer;
+    } else if (zeroCount == 1) {
+      for(int i = 0; i < n; i++) {
+        if(nums[i] != 0) {
+          answer[i] = 0;
+          continue;
+        }
+        answer[i] = product;
+      }
+    } else {
+      for(int i = 0; i < n; i++) {
+        answer[i] = product / nums[i];
+      }
+    }
+    return answer;
+  }
+}